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For prime $p$, suppose we have the base field $k = \mathbb{F}_p(x^p,y^p)$ and its extension $\ell = \mathbb{F}_p(x,y)$.

How could I prove that for some purely inseparable extension $K/k$ of degree $p$, $K\cong L$ where $\ell/L/k$?

etalic
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    So you know that $K=k(z)$ for some $z$, and $z^p\in k$? Does that help? – Jyrki Lahtonen Mar 23 '20 at 21:10
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    That's good. Ok, so $z^p$ is an element of $k$. Therefore there exists bivariate polynomials $a(T,U),b(T,U)\in\Bbb{F}_p[T,U]$ such that $$z^p=w=\frac{a(x^p,y^p)}{b(x^p,y^p)}.$$ By Freshman's Dream $a(x^p,y^p)=a(x,y)^p$. Ditto for $b$. Therefore both $z$ and $a(x,y)/b(x,y)\in\ell$ share the minimal polynomial $m(T)=T^p-w\in k[T]$. Therefore $k(z)$ is isomorphic to $k(a(x,y)/b(x,y))$. The latter is a subfield of $\ell$. – Jyrki Lahtonen Mar 24 '20 at 18:21

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$$(\Bbb{F}_p(x,y))^p = {\Bbb{F}_p}^ p(x^p,y^p)=\Bbb{F}_p(x^p,y^p)$$

That is to say $$\Bbb{F}_p(x,y)=(\Bbb{F}_p(x^p,y^p))^{1/p}$$

Given a field $F$ of characteristic $p$ then $F^{1/p}$ is the unique field extension generated by all the purely inseparable elements of degree $p$.

The latter fact follows from that $F^{1/p}$ is a field (isomorphic to $p$) and that for $a$ purely inseparable of degree $p$ over $F$ with minimal polynomial $f\in F[x]$ then $f$ has no other root ie. $f = (x-a)^p=x^p-a^p$ so that $a\in F^{1/p}$.

reuns
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